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PhysicsMediumMCQ2022 · 29 Jul Shift 2

Q17.Light enters from air into a given medium at an angle of 45° with interface of the air-medium surface. After refraction, the light ray is deviated through an angle of 15° from its original direction. The refractive index of the medium is : (1) 1 . 732 (2) 1 . 333 (3) 1 . 414 (4) 2 . 732

What This Question Tests

This question applies Snell's Law to determine the refractive index of a medium, given the angle of incidence and the angle of deviation after refraction.

Concepts Tested

Snell's LawAngle of incidenceAngle of deviationRefractive index

Formulas Used

n₁sinθ₁ = n₂sinθ₂

📚 NCERT Sections This Tests

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9.4 Figures 9.27(a) and (b) show refraction of a ray in air incident at 60° with the normal to a glass-air and water-air interface, respectively. Predict the angle of refraction in glass when the angle of incidence in water is 45° with the normal to a water-glass interface [Fig. 9.27(c)]. FIGURE 9.27

9.6A Prism Is Made Of Glass Of Unknown Refractive Index. A Parallel

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9.6 A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be 40°. What is the refractive index of the material of the prism? The refracting angle of the prism is 60°. If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light.

9.17(A) Sin I¢C = 1.44/1.68 Which Gives I¢C = 59°. Total Internal Reflection

Physics Class 12 · Chapter 9

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9.17 (a) sin i¢c = 1.44/1.68 which gives i¢c = 59°. Total internal reflection takes place when i > 59° or when r < rmax = 31°. Now, (sin i /sin r max max ) = 1.68 , which gives imax ~ 60°. Thus, all incident rays of angles in the range 0 < i < 60° will suffer total internal reflections in the pipe. (If the length of the pipe is finite, which it is in practice, there will be a lower limit on i determined by the ratio of the diameter to the length of the pipe.) (b) If there is no outer coating, i¢c = sin–1(1/1.68) = 36.5°. Now, i = 90° will have r = 36.5° and i¢ = 53.5° which is greater than i¢c. Thus, all incident rays (in the range 53.5° < i < 90°) will suffer total internal reflections.